Floating point -> string conversions

[John Fouhy]

Can someone explain these differences?

from math import pi
pi 3.1415926535897931 type(pi) <type 'float'> repr(pi) '3.1415926535897931' str(pi) '3.14159265359' '%f' % pi '3.141593' '%.16f' % pi

'3.1415926535897931'

Why do the string and default floating point representations get rounded?

(and is this documented anywhere?)

--
John.

[Alex Martelli]

John Fouhy <jf****@paradise.net.nz> wrote:
...

Why do the string and default floating point representations get rounded?

(and is this documented anywhere?)

Sure, right in the "Tutorial (start here)" for example.
http://www.python.org/dev/doc/devel/tut/node16.html
Alex

[John Fouhy]

al*****@yahoo.com (Alex Martelli) wrote in message news:<1gmxb89.r6ndtk17sltbgN%al*****@yahoo.com>...

John Fouhy <jf****@paradise.net.nz> wrote:
...

Why do the string and default floating point representations get rounded?

(and is this documented anywhere?)

Sure, right in the "Tutorial (start here)" for example.
http://www.python.org/dev/doc/devel/tut/node16.html
Alex

I knew I should have looked in the tutorial, rather than just the
"numeric types" section of the library reference :-/

Actually, that page doesn't mention that '%f' will give you a
different rounding again...

--
John.

[John Fouhy]

I have another related question...

pow(2, 31) 2147483648L '%d' % 2147483647.0 # python will convert to int '2147483647' '%d' % 2147483648.0 # too big for an int, so error. Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: int argument required '%d' % long(2147483648.0) # but yet, no trouble accepting a long. '2147483648' '%d' % int(2147483648.0) # and int() converts to long anyway

'2147483648'

Is this a bug?

(python 2.3.4)

--
John.

[Steve Holden]

John Fouhy wrote:

I have another related question...

pow(2, 31)
2147483648L
'%d' % 2147483647.0 # python will convert to int
'2147483647'
'%d' % 2147483648.0 # too big for an int, so error.
Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: int argument required
'%d' % long(2147483648.0) # but yet, no trouble accepting a long.
'2147483648'
'%d' % int(2147483648.0) # and int() converts to long anyway

'2147483648'

Is this a bug?

(python 2.3.4)

I don't think so. The fact of the matter is that a %d format token
explicitly expects an integral value.

The fact that %s coerces things makes us expect more of other format
tokens, but what would you expect %d to do with 2147483648.5?

regards
Steve
--
http://www.holdenweb.com
http://pydish.holdenweb.com
Holden Web LLC [phone]

[John Fouhy]

Steve Holden <st***@holdenweb.com> wrote in message news:<t9Skd.57$nj.56@lakeread01>...

The fact that %s coerces things makes us expect more of other format
tokens, but what would you expect %d to do with 2147483648.5?

I'd expect it to give me "2147483648", since that would be consistent
with its other behaviour :-)

(eg: eval('"%d" % 1.5') == '1')

I guess what annoys me is that python encourages you to not worry
about the difference between ints and longs (which is great) (unless
you're doing high performance computing), but then something like this
can cause code to stop working just because you've passed a border
which is almost invisible.

(of course, this means that one solution which I would be happy with
is for python to reject any non-integer/long arguments to "%d" :-) )

--
John.